Difference between revisions of "2015 AMC 10B Problems/Problem 13"

Latest revision as of 16:14, 2 August 2022

Problem

The line $12x+5y=60$ forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?

$\textbf{(A) } 20 \qquad\textbf{(B) } \dfrac{360}{17} \qquad\textbf{(C) } \dfrac{107}{5} \qquad\textbf{(D) } \dfrac{43}{2} \qquad\textbf{(E) } \dfrac{281}{13}$

Solution 1

We find the $x$ -intercepts and the $y$ -intercepts to find the intersections of the axes and the line. If $x=0$ , then $y=12$ . If $y$ is $0$ , then $x=5$ . Our three vertices are $(0,0)$ , $(5,0)$ , and $(0,12)$ . Two of our altitudes are $5$ and $12$ , and since it is a $5$ - $12$ - $13$ right triangle, the hypotenuse is $13$ . Since the area of the triangle is $30$ , so our final altitude is $\frac{30(2)}{13}=\frac{60}{13}$ . The sum of our altitudes is $\frac{60+156+65}{13}=\boxed{\textbf{(E)} \dfrac{281}{13}}$ . Note that there is no need to calculate the final answer after we know that the third altitude has length $\frac{60}{13}$ since $E$ is the only choice with a denominator of $13$ and $13$ is relatively prime to $5$ and $12$ .

Video Solution 1

https://youtu.be/zFHtdWEhriQ

~Education, the Study of Everything

Solution 2 (very similar to Solution 1)

Noticing that the line has coefficients $12$ and $5$ , we can suspect that we have a $5$ - $12$ - $13$ triangle on our hands. Because we want the sum of the altitudes, the third altitude that is not an axis must have a denominator of $13$ . Since the other altitudes are integers, we choose the option with $13$ as the denominator, namely $E$ .

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-We find the x-intercepts and the y-intercepts to find the intersections of the axes and the line. If <math>x=0</math>, then <math>y=12</math>. If <math>y</math> is <math>0</math>, then <math>x=5</math>. Our three vertices are <math>(0,0)</math>, <math>(5,0)</math>, and <math>(0,12)</math>. We multiply base by height and divide by <math>2</math> to get <math>\boxed{30}</math> as our answer.
+==Problem==
+The line <math>12x+5y=60</math> forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?
+<math>\textbf{(A) } 20 \qquad\textbf{(B) } \dfrac{360}{17} \qquad\textbf{(C) } \dfrac{107}{5} \qquad\textbf{(D) } \dfrac{43}{2} \qquad\textbf{(E) } \dfrac{281}{13} </math>
+==Solution 1==
+We find the <math>x</math>-intercepts and the <math>y</math>-intercepts to find the intersections of the axes and the line. If <math>x=0</math>, then <math>y=12</math>. If <math>y</math> is <math>0</math>, then <math>x=5</math>. Our three vertices are <math>(0,0)</math>, <math>(5,0)</math>, and <math>(0,12)</math>. Two of our altitudes are <math>5</math> and <math>12</math>, and since it is a <math>5</math>-<math>12</math>-<math>13</math> right triangle, the hypotenuse is <math>13</math>. Since the area of the triangle is <math>30</math>, so our final altitude is <math>\frac{30(2)}{13}=\frac{60}{13}</math>. The sum of our altitudes is <math>\frac{60+156+65}{13}=\boxed{\textbf{(E)} \dfrac{281}{13}}</math>. Note that there is no need to calculate the final answer after we know that the third altitude has length <math>\frac{60}{13}</math> since <math>E</math> is the only choice with a denominator of <math>13</math> and <math>13</math> is relatively prime to <math>5</math> and <math>12</math>.
+==Video Solution 1==
+https://youtu.be/zFHtdWEhriQ
+~Education, the Study of Everything
+==Solution 2 (very similar to Solution 1)==
+Noticing that the line has coefficients <math>12</math> and <math>5</math>, we can suspect that we have a <math>5</math>-<math>12</math>-<math>13</math> triangle on our hands. Because we want the sum of the altitudes, the third altitude that is not an axis must have a denominator of <math>13</math>. Since the other altitudes are integers, we choose the option with <math>13</math> as the denominator, namely <math>E</math>.
+==See Also==
+{{AMC10 box|year=2015|ab=B|num-b=12|num-a=14}}
+{{MAA Notice}}
+[[Category: Introductory Geometry Problems]]

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